Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A $0.5 \text{ kg}$ ball moving with a speed of $12 \text{ m/s}$ strikes a hard wall at an angle of $30^{\circ}$ with the wall. It is reflected with the same speed and at the same angle. If the ball is in contact with the wall for $0.25 \text{ s}$ , the average force acting on the wall is:

48 N

24 N

12 N

96 N

Step-by-Step Solution

Identify the Angle: The problem states the ball strikes at an angle of $30^{\circ}$ with the wall. This corresponds to a glancing angle. The angle with the normal (perpendicular to the wall) is $\theta = 90^{\circ} - 30^{\circ} = 60^{\circ}$ . Alternatively, one can resolve components directly using the angle with the wall.
Change in Momentum (Impulse): The force exerted by the wall changes the momentum of the ball only in the direction perpendicular to the wall.

Component perpendicular to the wall: $v_{\perp} = v \sin(30^{\circ})$ (using angle with wall) or $v \cos(60^{\circ})$ (using angle with normal).
Calculation of Impulse ( $\Delta p$ ): $\Delta p = m(v_{\perp, final} - v_{\perp, initial})$ $\Delta p = m(v \sin 30^{\circ} - (-v \sin 30^{\circ})) = 2mv \sin 30^{\circ}$ Substituting values: $\Delta p = 2 \times 0.5 \times 12 \times 0.5 = 6 \text{ kg m/s}$ .

Calculate Average Force: According to Newton's Second Law: $F_{avg} = \frac{\Delta p}{\Delta t}$ $F_{avg} = \frac{6}{0.25} = 24 \text{ N}$ (Reference: NCERT Class 11, Physics Part I, Chapter 5, Section 5.7 regarding Impulse).

Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started